All questions of Multiple Integrals for Mathematics Exam

To find the value of the integral, we need to set up the integral using the given triangle.

First, let's find the equation of the line AB. The slope of AB is (0-0)/(1-0) = 0, and since the y-coordinate of both points A and B is 0, the equation of the line AB is y = 0.

Next, let's find the equation of the line AC. The slope of AC is (1-0)/(0-0) = undefined, and since the x-coordinate of both points A and C is 0, the equation of the line AC is x = 0.

Now, we can set up the integral using the equation of the line AB and the equation of the line AC.

Since the triangle ABC is bounded by the x-axis, the y-axis, and the line AB, we can set up the integral as follows:

∫[y = 0 to y = 1] ∫[x = 0 to x = y] dx dy

Integrating the inner integral with respect to x, we have:

∫[y = 0 to y = 1] [x] [x = 0 to x = y] dy
= ∫[y = 0 to y = 1] y^2 dy

Integrating the outer integral with respect to y, we have:

∫[y = 0 to y = 1] (1/3) y^3 [y = 0 to y = 1]
= (1/3) [1^3 - 0^3]
= 1/3

Therefore, the value of the integral is 1/3.

Let's consider the change of variables from (x, y) to (u, v), given by x(u, v) = uv and y(u, v) = v/u.

To find the Jacobian of this transformation, we need to find the partial derivatives of x with respect to u and v, and the partial derivatives of y with respect to u and v:

∂x/∂u = v
∂x/∂v = u
∂y/∂u = -v/u^2
∂y/∂v = 1/u

The Jacobian of the transformation is then given by the determinant of the matrix:

J = |∂x/∂u ∂x/∂v|
|∂y/∂u ∂y/∂v|

J = |v u|
|-v/u^2 1/u|

The determinant of this matrix is J = v/u + v/u = 2v/u.

Now, to perform the change of variables in the double integral, we substitute x = uv and y = v/u into the integrand f(x, y):

f(x, y) → f(uv, v/u)

Finally, we need to account for the change in area element when transforming from (x, y) to (u, v). The area element dA in (x, y) coordinates is given by dA = dx dy, and in (u, v) coordinates it is given by dA' = |J| du dv.

Thus, the double integral with the change of variables is:

∫∫ f(uv, v/u) |J| du dv

To find the area common to the two circles, we need to determine the region where the two circles overlap.

The equation of the first circle is r = a, which represents a circle with radius a centered at the origin.

The equation of the second circle is r = 2a cos(θ), where θ is the angle measured from the positive x-axis.

To find the region of overlap, we need to find the values of θ where the two equations are equal:

a = 2a cos(θ)

Dividing both sides by a:
1 = 2 cos(θ)

Dividing by 2:
1/2 = cos(θ)

This equation is true when θ = π/3 and θ = 5π/3.

So, the region of overlap occurs between θ = π/3 and θ = 5π/3.

To find the area of this region, we integrate the equation for the second circle from θ = π/3 to θ = 5π/3:

A = ∫[π/3, 5π/3] (1/2) * (2a cos(θ))^2 dθ

Simplifying,
A = (1/2) * 4a^2 * ∫[π/3, 5π/3] cos^2(θ) dθ

Using the identity cos^2(θ) = (1 + cos(2θ))/2,
A = (1/2) * 4a^2 * ∫[π/3, 5π/3] (1 + cos(2θ))/2 dθ

Simplifying further,
A = 2a^2 * ∫[π/3, 5π/3] (1 + cos(2θ))/2 dθ

Integrating,
A = 2a^2 * [(θ/2) + (sin(2θ)/4)] evaluated from θ = π/3 to θ = 5π/3

Evaluating at the limits,
A = 2a^2 * [(5π/3 - π/3)/2 + (sin(10π/3)/4 - sin(2π/3)/4)]

Simplifying,
A = 2a^2 * [(4π/3)/2 + (sin(10π/3) - sin(2π/3))/4]

A = 2a^2 * [(2π/3) + (sin(10π/3) - sin(2π/3))/4]

This is the area common to the circles r = a and r = 2a cos.

It seems like the equation is incomplete. Could you please provide the missing information?